# Proving that convergence of a sequence of functions is not uniform

I have the following sequence of functions;

$f_n(x) = \begin{cases} 4n^2x & \text{if$0 \leq x \leq \frac{1}{2n} $} \\ -4n^2x+4n& \text{if$\frac{1}{2n} \leq x \leq \frac{1}{n} $} \\ 0 & \text{otherwise} \end{cases}$.

I want to show that $\lim_{n\rightarrow \infty} f_n(x) = 0$, and that its convergence on $[0,1]$ is not uniform. I am given the result that the limit of a sequence of riemann integrable functions need not be riemann integrable.

• What have you tried so far? It is almost certainly going to be a case of assuming that it is uniformly convergent, and then working with the definition of uniform convergence to derive a contradiction. Please demonstrate some working on your part. – EHH Mar 16 '16 at 12:59
• I've shown that the function is symmetric about 1/2n and that it is riemann integrable on [0,1], since the measure of it's discontinuities is zero. I understand how to use the definition of uniform convergence to prove convergence, but not to negate it. – Dr. John A Zoidberg Mar 16 '16 at 13:02

Note that as $n \to \infty$, the interval $\left[ 0, \dfrac 1 n \right]$ on which the graph of $f_n$ looks like a tent shrinks down to a point, while its complementary $\left[ \dfrac 1 n, 1 \right]$ on which $f_n$ is $0$ expands to cover the whole of $[0,1]$, therefore $f = \lim f_n = 0$ (the constant function $0$).

In general, to say that $f_n$ converges uniformly to $f$ on some set $A$ means that $\lim \limits _n \sup \limits _{x \in A} |f_n (x) - f(x)| = 0$. In our case this gives $\lim \limits _n \sup \limits _{x \in [0,1]} f_n (x) = 0$ (because $f = 0$ and $f_n \ge 0$). But what is $\sup \limits _{x \in [0,1]} f_n (x)$? Well, it is the value at the "top of the tent", which is $2n$ (and it is reached in the point $x = \dfrac 1 {2n}$, but this is not important here). Since $2n \not\to 0$, then the convergence is not uniform. (Note that this has nothing to do with integrability, the presence of which in this problem seems to me to be a "red herring".)